Door: http://poj.org/problem? Id = 1442
Black Box
Time limit:1000 ms |
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Memory limit:10000 K |
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Description
Our black box represents a primitive database. it can save an integer array and has a special I variable. at the initial moment black box is empty and I equals 0. this black box processes a sequence of commands (transactions ). there are two types of transactions:
Add (x): Put element x into black box;
Get: Increase I by 1 and give an I-Minimum Out Of All integers containing in the black box. keep in mind that I-minimum is a number located at I-th place after Black Box elements sorting by non-descending.
Let us examine a possible sequence of 11 transactions:
Example 1
N Transaction i Black Box contents after transaction Answer
(elements are arranged by non-descending)
1 ADD(3) 0 3
2 GET 1 3 3
3 ADD(1) 1 1, 3
4 GET 2 1, 3 3
5 ADD(-4) 2 -4, 1, 3
6 ADD(2) 2 -4, 1, 2, 3
7 ADD(8) 2 -4, 1, 2, 3, 8
8 ADD(-1000) 2 -1000, -4, 1, 2, 3, 8
9 GET 3 -1000, -4, 1, 2, 3, 8 1
10 GET 4 -1000, -4, 1, 2, 3, 8 2
11 ADD(2) 4 -1000, -4, 1, 2, 2, 3, 8
It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and get transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer Arrays:
1. A (1), a (2 ),..., A (m): a sequence of elements which are being encoded into black box. A values are integers not exceeding 2 000 000 000 by their absolute value, m <= 30000. for the example we have A = (3, 1,-4, 2, 8,-1000, 2 ).
2. U (1), u (2 ),..., U (n): a sequence setting a number of elements which are being encoded into black box at the moment of first, second ,... and N-transaction get. for the example we have u = (1, 2, 6, 6 ).
The Black Box algorithm supposes that natural number sequence U (1), u (2 ),..., U (n) is sorted in non-descending order, n <= m and for each P (1 <= P <= N) an inequality P <= U (P) <= m is valid. it follows from the fact that for the p-element of our U sequence we perform a get transaction giving p-minimum number from our A (1 ), A (2 ),..., A (U (p) sequence.
Input
Input contains (in given order): m, n, A (1), a (2 ),..., A (M), u (1), u (2 ),..., U (n ). all numbers are divided by spaces and (or) carriage return characters.
Output
Write to the output black box answers sequence for a given sequence of transactions, one number each line.
Sample Input
7 43 1 -4 2 8 -1000 21 2 6 6
Sample output
3312
Source
Northeastern Europe 1996 second treap .. In general, treap is very easy to write. I learned treap to prevent some problem cards from being splay .. It seems like Qaq... there is no way to do this .. Codes:
1 #include<set> 2 #include<ctime> 3 #include<queue> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std;10 const int N = 100100;11 #define L(i) (T[i].s[0])12 #define R(i) (T[i].s[1])13 #define For(i,n) for(int i=1;i<=n;i++)14 #define Rep(i,l,r) for(int i=l;i<=r;i++)15 16 struct treap{17 int size,s[2],v,pri;18 void Sets(int x,int y){19 size = 1;v = x;pri = y;20 }21 }T[N];22 23 int n,m,A[N],size,Lim,now,level = 0;24 int tot,root;25 int read(){26 char ch = getchar(); int num = 0 , q = 1;27 while(ch>‘9‘||ch<‘0‘){28 if(ch==‘-‘) q = -1;29 ch = getchar();30 }31 while(ch>=‘0‘&&ch<=‘9‘){32 num = num * 10 + ch - ‘0‘;33 ch = getchar();34 }35 return num * q;36 }37 38 void Update(int i){39 T[i].size = T[L(i)].size + T[R(i)].size + 1;40 }41 42 void Rot(int &y,int f){43 int x = T[y].s[!f];44 T[y].s[!f] = T[x].s[f];45 T[x].s[f] = y;46 Update(y);Update(x);47 y = x;48 }49 50 void Insert(int &i,int val){51 if(!i){52 T[i=++tot].Sets(val,rand());53 return;54 }55 int f = T[i].v > val;56 Insert(T[i].s[!f],val);57 if(T[T[i].s[!f]].pri > T[i].pri) Rot(i,f);58 else Update(i);59 }60 61 int Rank(int i,int kth){62 if(T[L(i)].size + 1 == kth) return i;63 else if(T[L(i)].size >=kth) return Rank(L(i),kth);64 else return Rank(R(i),kth - T[L(i)].size - 1);65 }66 67 int main(){68 srand(time(NULL));69 n = read(); m = read();70 For(i,n) A[i] = read();71 For(i,m) {72 Lim = read();73 Rep(i,now+1,Lim) Insert(root,A[i]); now = Lim;74 level++;printf("%d\n",T[Rank(root,level)].v);75 }76 return 0;77 }